{"id":24943,"date":"2020-09-15T15:58:51","date_gmt":"2020-09-15T10:28:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=24943"},"modified":"2021-07-02T15:01:49","modified_gmt":"2021-07-02T09:31:49","slug":"kseeb-solutions-for-class-10-maths-chapter-1-ex-1-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-1-ex-1-4\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4<\/h2>\n

Question 1.
\nWhich term of the A.P: 121, 117, 113,……, is its first negative term?
\n[Hint: Find n for an<\/sub> < 0]
\nAnswer:
\nThe given AP is 121, 117, 113,…….
\na = 121, d = 117 – 121 = – 4
\nLet the nth term of the AP be the first negative term.
\n\u2234 an<\/sub> < 0
\na + ( n – 1 )d < 0
\n121 + (n – 1) (- 4) < 0
\n121 – 4n + 4 < 0
\n125 < 4n
\n4n > 125
\n\"Karnataka
\n\u2234 n = 32.
\nHence, 32nd<\/sup>\u00a0term of the given AP is the first negative term.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nThe sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
\nAnswer:
\nLet the first term be \u2018a\u2019 and common difference is ‘d’
\na3<\/sub> + a7<\/sub> = 6.
\na + 2d + a + 6d = 6
\n2a + 8d = 6 divide by 2
\na + 4d = 3
\na = 3 – 4d \u2192 (1)
\na3<\/sub> \u00d7 a7<\/sub> = 8
\n(a + 2d) (a + 6d) = 8
\n(3 – 4d + 2d) (3 – 4d + 6d) = 8
\n(3 – 2d) (3 + 2d) = 8
\n\"Karnataka
\n\"Karnataka
\n\"Karnataka<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nA ladder has rungs 25 cm apart. (see Fig.1.7 ). The rungs decrease uniformly \u00a1n length form 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2 \\(\\frac{1}{2}\\) m apart, what is the length of the wood required for the rungs?
\n[Hint: Number of rungs \\(\\frac{250}{25}\\) + 1].<\/p>\n

\"Karnataka
\nAnswer:
\nDistance between top and bottom rung
\n\"Karnataka
\nDistance between every two rungs
\n= 25 cm [given]
\nNumber of rungs (n) = + 1 for top most rung]
\n= 10 + 1 = 11
\n\u2234 n = 11
\nBecause the rungs decrease uniformly in length from 45 cm at the bottom to 25cm at the top
\n\"Karnataka<\/p>\n

Question 4.
\nThe houses of a row are numbered consecutively from Ito 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
\n[Hint: Sx-1 <\/sub>= S49<\/sub> – Sx<\/sub>]
\nAnswer:
\nThe numbers of houses are 1, 2, 3, 4, ……… 49
\na = 1 & d = 1
\nsum of n terms of a AP
\nS = \\(\\frac{n}{2}\\)[2a + (n – 1)d]
\nSum of number of houses preceding xth <\/sup>house is equal to Sx-1<\/sub> i.e.,<\/p>\n

\"Karnataka
\nSum of numbers of houses following xth<\/sup> house is equal to S49<\/sub> – Sx<\/sub>.
\n\"Karnataka<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nA small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \\(\\frac{1}{4}\\) m and a tread of \\(\\frac{1}{2}\\) m (see Fig 5.8). Calculate the total volume of concrete required to build the terrace.
\nHint: Volume of concrete required to build the first step = \\(\\frac{1}{4} \\times \\frac{1}{2} \\times 50 \\mathrm{m}^{3}\\)
\n\"Karnataka
\nAnswer:
\nVolume of concrete required to build the first step length \u00d7 breadth \u00d7 Height.\"Karnataka
\n\u2234 Total volume of concrete required to build the terrace of 15 steps
\n\"Karnataka<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Question 1. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/24943"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=24943"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/24943\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=24943"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=24943"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=24943"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}